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Chapter 14: Problem 86

Consider a car fitted with a catalytic converter. The first 10 min or so after it is started are the most polluting. Why?

Short Answer

Expert verified
The first 10 minutes after a car with a catalytic converter is started are the most polluting because the catalytic converter, which reduces harmful emissions, needs to heat up to a certain temperature to effectively facilitate the necessary reactions to reduce pollution. During this initial period, it is below its operating temperature and hence, not working optimally resulting in higher pollution.

Step by step solution

01

Understanding catalytic converters

A catalytic converter is a device installed in vehicles to reduce the emissions of harmful pollutants. It does this by facilitating chemical reactions that convert these harmful pollutants into less harmful substances before they are emitted from the vehicle's exhaust system.
02

The role of temperature in the working of catalytic converters

For a catalytic converter to work effectively, it needs to heat up to a certain temperature. The catalysts incorporated into the converter promote reduction and oxidation reactions that convert pollutants into less harmful substances. These reactions are temperature-dependent and occur optimally at high temperatures.
03

Why initial minutes are most polluting

When a car that has been parked for a while is started, the catalytic converter is at ambient temperature and hence, is below its optimal operating temperature. This implies it cannot effectively facilitate the necessary reactions to reduce pollution. Therefore, during the initial minutes, until the catalytic converter reaches its optimal temperature as a result of the engine's heat, the car exhaust emits more pollutants making these first few minutes most polluting.

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Most popular questions from this chapter

The reaction \(2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C}\) is first order with respect to \(\mathrm{A}\) and \(\mathrm{B}\). When the initial concentrations are \([\mathrm{A}]=1.6 \times 10^{-2} M\) and \([\mathrm{B}]=2.4 \times 10^{-3} M,\) the rate is \(4.1 \times 10^{-4} M / \mathrm{s} .\) Calculate the rate constant of the reaction.

Explain why most metals used in catalysis are transition metals.

Consider the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of \(0.074 M / \mathrm{s}\). (a) At what rate is ammonia being formed? (b) At what rate is molecular nitrogen reacting?

Explain what is meant by the rate law of a reaction.

To carry out metabolism, oxygen is taken up by hemoglobin (Hb) to form oxyhemoglobin \(\left(\mathrm{HbO}_{2}\right)\) according to the simplified equation $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times\) \(10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). (The reaction is first order in \(\mathrm{Hb}\) and \(\mathrm{O}_{2} .\) ) For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} M\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} \mathrm{M} / \mathrm{s}\) during exercise to meet the demand of increased metabolism rate. Assuming the Hb concentration to remain the same, what must be the oxygen concentration to sustain this rate of \(\mathrm{HbO}_{2}\) formation?
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